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Suppose a $T_4$ space $X$ satisfies a property that every continuous map $f$ from $X$ to any Hausdorff space $Y$ is a closed map. Show that $X$ is compact.

What I am thinking is to let $\mathcal{U}$ be an open cover of $X$ and find a way to choose finite open sets from $\mathcal{U}$ that can cover $X$ again. Do we need to define a suitable continuous map by ourselves and use the fact that it is a closed map to choose suitable open sets from $\mathcal{U}$? I have tried to define a continuous map $f \colon X \to Y$ as mapping different open sets from $\mathcal{U}$ to different points in $Y$. However, different open sets from $\mathcal{U}$ may intersect with each other and hence $f$ may not be well-defined. Apart form that, I have no idea about how to choose suitable $f$ and $Y$. Can someone help me please?

Edit 1: I am trying to apply the Urysohn Lemma or the Tietz Extension Theorem due to the normality of $X$. But still have no idea.

Edit 2: Finally found a way to prove without using compactification. I will post the proof when I’m free. Hint: Use Urysohn Lemma and Tychonoff Theorem.

ansonngg
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2 Answers2

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Any such $X$ has a compactification $(\gamma X, e)$ in which $e: X \to \gamma X$ is an embedding with a dense image (this uses $T_{3\frac12}$). Then the assumption says that $e$ must be a closed map, so $\gamma X = e[X]$ and $X$ is compact.

Henno Brandsma
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  • Can you briefly explain why $X$ must have a compactification? Because I haven't learnt anything about compactification but just a little of quotient space (I don't know but I think they are related after searching on Wiki). – ansonngg Mar 23 '21 at 09:50
  • @ansonngg Every Tychonoff space embeds into $[0,1]^I$ for some $I$, and we can take its closure there. This is classical. – Henno Brandsma Mar 23 '21 at 10:23
  • I think I need some time to digest. Is it possible to not use compactification to deal with the question? Since it's still not covered in my lectures and I think my professor doesn't expect us to use such concept. But anyway, thanks so much! – ansonngg Mar 23 '21 at 10:51
  • @ansonngg maybe the $T_4$ slight overkill is a hint to use the Tietze extension theorem or Urosohn's lemma? – Henno Brandsma Mar 23 '21 at 10:56
  • I've tried to apply these two theorems but still have no idea how to use them. – ansonngg Mar 23 '21 at 12:09
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This is essentially Henno’s argument, but without relying on previous knowledge of compactifications. Added: And it may be the argument that you found. (I had this tab sitting open and didn’t notice your second edit before posting.)

Let $\mathscr{B}$ be a base for $X$, and for each $B\in\mathscr{B}$ let $x_B\in B$; by normality there is a continuous $f_B:X\to[0,1]$ such that $f_B(x_B)=1$ and $f_B[X\setminus B]=\{0\}$. For each $B\in\mathscr{B}$ let $I_B$ be a copy of $[0,1]$, and let $Y=\prod_{B\in\mathscr{B}}I_B$ with the product topology. Finally, let

$$h:X\to Y:x\mapsto\langle f_B(x):B\in\mathscr{B}\rangle\,.$$

  • Show that $h^{-1}[\{y\in Y:y_B\in U\}]$ is open in $X$ for each $B\in\mathscr{B}$ and open $U\subseteq I_B$. Conclude that $h$ is continuous and therefore by hypothesis closed.
  • Use the fact that $\mathscr{B}$ is a base for $X$ to show that $h$ is injective.
  • Conclude that $h$ is a homeomorphism of $X$ onto $h[X]$.

Now $h[X]$ is a closed subset of the compact space $Y$, so $h[X]$ is compact and therefore so is $X$.

Brian M. Scott
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  • Originally, I planned to post my answer today but just found your answer and yours is pretty similar to mine. So I decided to choose yours to be the correct answer of this question. Anyway, thank you so much! – ansonngg Apr 03 '21 at 11:35
  • @ansonngg: You’re welcome! (And I’m delighted that you were able to solve it yourself.) – Brian M. Scott Apr 03 '21 at 18:12